What is the solution for 9y^{''}-12y^'+4y=0,y(0)=e^{(2x)/3} ?

The solution for 9y^{''}-12y^'+4y=0,y(0)=e^{(2x)/3} is y=c_{1}e^{(2x)/3}+c_{2}e^{(2x)/3}x

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9y^{''}-12y^'+4y=0,y(0)=e^{(2x)/3}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2 x}{3}}

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{\frac{2 x}{3}} x

Solution steps

9 y^{''} (x) - 12 y^{'} (x) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{\frac{2 x}{3}} x

y = c_{1} e^{\frac{2 x}{3}} + c_{2} e^{\frac{2 x}{3}} x

y^{^{''''}} + 3 y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (1) = 7

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 6

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

4 y^{^{''}} - 24 y^{^{'}} + 37 y = 0, y (0) = - 9

What is the solution for 9y^{''}-12y^'+4y=0,y(0)=e^{(2x)/3} ?
The solution for 9y^{''}-12y^'+4y=0,y(0)=e^{(2x)/3} is y=c_{1}e^{(2x)/3}+c_{2}e^{(2x)/3}x